# Optimization

Copyright © 2003–2017 by Stan Brown

Copyright © 2003–2017 by Stan Brown

**Summary:**
One of the main applications of the derivative is
**optimization problems** — finding the value of one
quantity that will make another quantity reach its largest or smallest
value, as required. Here’s an overview of the solution techniques.

- Write the
**primary equation**, the formula for the quantity to be optimized. The quantity to be optimized is the**dependent variable**, and the other variables are**independent variables**. Write down whether the dependent variable is to be maximized or minimized. - Before you can proceed, the primary equation must contain only one
independent variable and the dependent variable. If you have
extra variables, as usually you do, you must construct one or more
**secondary equations**that involve the independent variables in the primary equation. (Usually a problem constraint will lead to a secondary equation.) Substitute the secondary equations in the primary equation to eliminate the extra variables. - Write down the
**feasible domain**, the values of the remaining independent variable that make sense in the problem. Usually this information will come from the problem or your common-sense knowledge (such as no negative areas). - Find the desired maximum or minimum. (See summary below.)
- Write the answer to the problem. Make sure you have answered the
**actual question**!

A standard US can of soda (or pop, depending on where you live) holds 12 fluid ounces or 355 ml. Find the dimensions of a cylindrical can that will use the least amount of aluminum.

**Solution**:
The dependent variable is the amount of aluminum. Essentially,
you must minimize the surface area of the cylinder. Write the primary
equation: the surface area is the area of the two ends (each
πr²) plus the area of the side or lateral area

to minimize: A = 2πr² + 2πrh

The primary equation contains two independent variables, r and h. Can you relate them in some way? Yes, the problem constraint is that the volume equals 355 ml (or 355 cm³). This gives the secondary equation:

V = πr²h = 355

Since h occurs once in the primary equation, and as a linear term, it will be easy to eliminate. Solve the secondary equation for h:

h = 355 / πr²

Substitute in the primary equation:

A = 2πr² + 2πr·355/πr² ⇒ A = 2πr² + 710/r

There are no obvious constraints on the feasible domain, except that r and h must both be positive. Since the feasible domain has no endpoints, you need not check whether they are minima.

To try to find maxima or minima, differentiate:

dA/dr = 4πr − 710/r²

d²A/dr² = 4π + 1420/r³

Find critical numbers where dA/dr = 0 or does not exist. The derivative does not exist at r = 0; however you can disregard that because r = 0 is outside the feasible domain.

4πr − 710/r² = 0 ⇒ r = cuberoot(710/4π) ≈ 3.84 cm

Is this a minimum, a maximum, or neither? Since d²A/dr² is positive for all positive r, the critical point r = 3.84 cm must be a minimum.

The problem asked for the *dimensions* of the can with
lowest surface area, which means that you also need the height. To find it,
substitute r = 3.84 in the secondary equation and get
h ≈ 7.67 cm.

**Answer**: A cylindrical can with volume
355 ml will use the least aluminum if its radius is about
3.84 cm and its height is about 7.67 cm.

Check: V = πr²h = π(3.84²)(7.67) = 355.3 cm³, the same as the required volume give or take a little rounding difference.

Because a real soda can is not exactly cylindrical, you can’t expect perfect agreement with these figures. I measured about 3.2 cm radius and 12.7 cm height on a can of Barq’s root beer, including the top and bottom extensions.

By the way, you may have noticed that the radius is about
½ the height. In fact, you can prove that *the cylinder of a
given fixed volume with the lowest surface area will always have
r = h/2.* Instead of setting V = 355 keep V as
a letter and treat it as a constant. You will have
r = cuberoot(V/2π) and substituting
V = πr²h gives r = h/2.

**Differentiate**with respect to the independent variable. You will need both the first and the second derivatives.- Find the
**critical numbers**, where f′ is 0 or does not exist. - For
*each*critical number (call it c), evaluate the second derivative f″(c) to find whether you have**a maximum, a minimum, or neither**, as follows:- If f″(c) is negative, you have a hilltop and c gives you a maximum.
- If f″(c) is positive, you have a valley and c gives you a minimum.
- If f″(c) is 0 or does not exist, you must use the
First Derivative Test, which is:
- If f′(x) goes from positive to negative at x=c, you have a hilltop (maximum) at x=c.
- If f′(x) goes from negative to positive at x=c, you have a valley (minimum) at x=c.
- If f′(x) has the same sign on both sides of x=c, you have neither a maximum nor a minimum at x=c.

- You must also evaluate the original function (primary equation) for
**each endpoint of the feasible domain**, if it has endpoints. Compare to the function values at the optimum points from step 3.

**1 Jan 2014**: Remove references to separate page on relative maxima and minima, which was withdrawn today.- (intervening changes suppressed)
**22 Jul 2003**: New article.

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